Phys 306 Lectures 1 Waves Media 2011

7/31/2019 Phys 306 Lectures 1 Waves Media 2011

1/25

PHYS 306 - Optical Physics

Lecture Notes

2011

A/Prof. Mike Steel

Room E7A 207 [email protected]

Original notes prepared by A/Prof. David Coutts

Copyright Macquarie University 2011


2/25


3/25

3

Extension:

Give physical explanations for Maxwells Equations. For example, what does S

d B s

mean both mathematically and physically? Why should it equal zero?

Check for yourself thatc = (00)-1/2

using the known values of c, 0and0.

1.2 Wave Equations in empty space

Take the curl oft

=

BE , and use = grad div - 2 (note E = 0)

( )2

2

0 0 2.

t t

= = =

EE E B (1.2)

This gives us a second order differential wave equation:2

2

0 0 2t

=

E

E (1.3)

where the speed of the wave is0 0

1 /c = = (00)-1/2

.

The parameters 0 and 0 and the speed of light had all been measured by the time Maxwell workedall this out, and he found that the waves which were solutions to his wave equation would travel

with the speed of light. ie.

Thus in a flash of inspiration, he realised that light was an electromagnetic wave. (Of course, were

bending the truth here. Maxwell didnt have the advantage of vector notation and he certainly

didnt use SI. As we saw in PHYS301, in SI, c,0 and 0 are now all defined quantities, and it the

strength of the electromagnetic interaction is hiding in the definition of the metre.)

A similar procedure for B yields a wave equation of the same form as (1.3) for B.

1.3 Electromagnetic Plane Waves

Solutions to the wave equation will take the general form of:

( ) ( )( ) ( )

0

0

, exp

, exp

x y i kz t

x y i kz t

= =

E E

B B(1.4)

wherez is the direction of propagation.

We are going to consider plane wave solutions - what do Maxwells Equations say about the

directions ofB and E?

A plane wave means that there should be no variation ofB and E in thex andy directions:

0x y

= =

E E(1.5)

but from Gausss Law in vacuum, we have:

0yx z

EE E

x y z

= + + =

E (1.6)


4/25


5/25

5

1.4 Maxwells Equations Inside Media (Griffiths4.14.4.1, 7.3)

1.4.1 Electric Fields Inside Media

Remember that an electric field within a polarisable medium will induce a charge separation in

each atom in the medium ( = dipole ), ie. it polarises the medium to give a net dipolemoment. (The polarisation can also arise from the applied field causing a rotation and alignment of

the permanent dipole moments of polar molecules.)

Uniform Electric Field

Atoms in a medium, no E field Atoms in an applied uniform E field

The net effect of the external field is to induce a polarisation, which at any point can be described

by the dipole moment per unit volume P. So for a uniform distribution of dipoles each with dipole

moment p and number density n, we have n=P p .

Each dipole produces a field:

If we now reverse the problem, and ask what electric field E will be produced by a given

polarisation P?

Our study of the potential in PHYS301 has given us the tools to work this out. Recall these three

results for the scalar potential:

( )( )

( )( )

( )( )

dip 3

0

0

0

'1potential at due to dipole at

4 '

1potential at due to surface charge on S

4 '

1potential at due to charge in vo

lume V4 '

S

V

V

V da

V d

=

=

=

p r rr r r

r r

rr r

r r

rr r

r r

(1.10)

where the source point is r and observation point is r. To find the total potential from a

distribution of charge, we sum over all the individual charges with the first of Eqs. (1.10), and since

the dipole strength varies slowly on the atomic scale, we convert the sum to an integral to give:

E > 0E = 0


6/25

6

( )

( ) ( )

( ) ( )

( )

( )( )

0

0

0

0

0

3

3

3

2

'

'

'

'

''

1 )

1

4

/1

4

14

1 1, (using a result from 301:

4

1 1(int

'

=

' 'egr

4

i

i

i i

i

i

V

V

S V

V

vv

d

d

d d

=

=

=

=

r r

r r

r r

r r

r rr r

x

r r

p

p

P r

P r

P rP

x x

r r rr a

r

( )( )

0

ating by parts)

1 1

4 ' '=

S V

da d

n r r

P rP r

r r

(1.11)

The resulting potential has two terms, one that looks like the potential of a surface charge

( ) ( ) b =r P r n , and one that looks the potential of a volume charge density b = P . The

suffix b indicates that these are boundcharges stuck to atomstheyre not free to move about the

medium.

The surface charge term shows up at the edges of a uniform medium, where the charge isnt

cancelled by neighbouring opposite charges, (unless the orientation of the dipoles is perpendicular

to the surface).

Extensionwhat is the significance of the dot product with the surface normal in

b

= P n ? What would happen if it were not present, eg. if we hadb

= P ? Hint:

consider the edge of a region of charge for different directions of the applied field.

The bound charge volume density only shows up if the electric field is non-uniform, or if the

medium is non-uniform. In that case, the polarisation will not be uniform, ie. PP(x,y,z) and wecan get a non-vanishing divergence in

b .

We have bound charges within the medium, giving rise to a volume charge density which produces

an electric field.

The medium may also have free charges imbedded within it resulting in a free charge densityf

from charges that we put there, in addition to the induced polarisation charge densityb.


7/25

7

The total (volume) charge density that goes in to Gauss law becomes

b f = + .

And Gausss law now reads:

0 total b f f = = + = +E P .

Note that E is now the total fieldincluding that portion generated by the polarisation. Now, while

we can specify thefree chargef, we usually dont know the bound charge until the problem issolved.

However, we can combine the two div terms to give

( )0 ,f + =E P

and from this we define Das

0 , +D E P

where Dis the electric displacement. Note that the meaning ofDis a little subtle, as compared to

the way that Pis the dipole moment per unit volume, and E determines the force on a charge. First

and foremost, D is a device which allows us to perform calculations more easily, but it does play an

important role in the development of electromagnetic and especially optical theory. (Its also

important to remember that properties ofE, do not necessarily apply to D. For example, D is in

general not curl-free, and therefore there is no potential for D.)

Gausss law now reads:

encl,

fd Q == DD s (1.12)

where Qencl. is the totalfree charge enclosed within the volume. Remember, we can control freecharge, so finding Dis easier than finding E.

To close the system of equations, we need a connection between the polarisation and the electric

field. If the applied electric field is not too strong, we can assume a linear response for which the

induced polarisation is proportional to the electric field:

0 e =P E ,

wheree is the electric susceptibility(it is dimensionless). We now have

( )

0

0 0

0 1 e

e

= +

= +

= +

=

E

D E P

E

E

E

and thus both P and D are proportional to E, with

=D E ,

where ( )0 1 e = + is thepermittivity of the material, (cf. 0 =permittivity of free space). It is

also common to define r= 1+e as the relative permittivity of the medium, so D = 0rE. At

optical frequencies, typical dielectrics have values ofr

in the range [1,15].


8/25

8

It is important to note that the assumption of linear response is only an approximation and is only

valid at modest electric field strengths. Later in the course, we will begin to consider the

fascinating field ofnonlinearoptics, where these assumptions break down.

A point of terminology: we commonly speak ofnonlinear dielectrics and occasionally of linear

dielectrics, but no medium is always linear. Pushed sufficiently hard, any system will behave

nonlinearly, but the threshold electric fields at which this occurs varies markedly from one material

to another. Usually linear dielectric indicates that the system is being operated in a regime wherenonlinear effects may be neglected. Nonlinear dielectric may indicate the opposite regime, but is

also used to refer to materials whose nonlinear properties are desirable somehow. What those

desirable properties might be, well get to later.

1.4.2 Magnetic fields in media

Real materials can have a magnetisation which can be thought of as the field produced by bound

currents, as opposed to those currents over which we have control, calledfreecurrents. Thus we

can do a similar treatment as with the electric field where we considered bound and free charges.

The magnetic analog to the polarization is the magnetization M, defined such that

0/ = H B M . In linearmagnetic media, we assume that

=B H

where= 0(1+m), and m is the magnetic susceptibility. We can also write = 0r, with rknown as the relative permeability. Thus the auxiliary field His used in magnetostatics in an

analogous way to the electric displacement Din electrostatics. For most dielectrics,

0 and 1m r and the magnetic properties are indistinguishable from free space. Thus for our

purposes, we can consider H and B to be identical apart from a constant multiplicative factor. Note

that due to one of those unfortunate historical quirks, enters the theory upside-down as

compared to : its , but = =D E B H .

Within a medium, Maxwells equations now become:

encl Gauss' Law

0 0

Faraday's Law

Ampe

re's L a

w

f

S

S

f f

d Q

d

d d = -

t t

d I d = +t t

= =

= =

=

= +

D s D

B s B

B BE l E

D DH l s H J

s

Note that the names given to the four fields vary according to era, country of origin and personal

preference. Most people agree on electric fieldfor E, and electric displacementfor D, but opinion

varies for the magnetic quantities. Many authors call H the magnetic fieldand B the magnetic flux

density, rather than calling B the magnetic field. Most practising physicists just refer to the symbols

by letter, and stay out of the debate.


9/25

9

1.5 Wave Equations Inside a Medium (Griffiths 9.3)

Now lets consider wave equations inside a medium with no free charge and no free current.

Maxwells equations become:

0 0

= - =t t

= =

D B

B DE H

but if the medium is linear, then =D E and =B H , and if medium is homogeneous, so that

and are independent of position, the equations reduce to:

0 0

= - =t t

= =

E B

B EE B

which look just like the free space case except that we have replaced00 by .Obviously, in a medium, we can derive the same wave equations, but with a propagation speed of

0 01 /

r rv = , and correspondingly v = c/n where n is the refractive index.

As weve said, for most materials r= 1, so

1r e

n = = + .

1.5.1 Boundary Conditions in media (Griffiths 7.3.6)

Now we are equipped to deal with Maxwells equations at a boundary in media. Weve seen thesederivations a number of times in 202 & 301, so we wont belabour the point in class, but the details

are here for the interested...

Recall that the issue is that Maxwells equations or the wave equation, only apply in homogeneous

or smoothly varying regions where we can safely take derivatives. At sharp (or fairly sharp)

boundaries, we need connection formulae to join up the solutions in each region.

Consider Gauss law,encl d Q= D s

applied to the following boundary:

If the enclosed volume is small

enough, then Gauss law reduces to

( )1 2 fs s=D n D n where f is the free

surface charge density (which will be zero in our case).

2

12

1ds


10/25

10

Thus, for zero free surface charge, the components ofDperpendicularto the surface are continuous

across the boundary:

1 2 =D D

and if the medium is an isotropic dielectric (see later) we have

1 1 2 2

=E E

Using the second of Maxwells laws, remembering that magnetic monopoles (probably) dont exist,

we find by the same argument that:

1 2 =B B

Thus the components ofBperpendicularto the surface are continuous across the boundary.

Applying Faradays law to a very thin loop half embedded in the surface, we find1 2 / dt= E l E l B s

where d da=s n is the directed area of the loop. But for our very thin loop, da = 0, so we have

1 2=E E

,

ie. the components of the electric fieldparallelto the interface are continuous across the boundary.

Finally using Amperes law and assuming no surface currents on our dielectric, we obtain :

1 2=H H

,

ie. the components ofHparallelto the interface are continuous across the boundary.

To summarise, the boundary conditions for nonmagnetic dielectrics are

1 2 1 2

1 2 1 2

D D B B = =

= =H H E E

(1.13)

and in terms ofE and B only,

1 1 2 2 1 2

1 2

.

E E = =

=

E E

B B

(1.14)

l

E2

E1

2

1


11/25


12/25

12

Now r can be any point in the pl

(k- kr) and (k kt) must be norm

same plane normal to the reflecti

Extensionshow this diagra

Now consider r=k r k r . Its cimplies sin sini r rkr k r = . Sin

giving us theLaw of reflection:

Angle of incidence = An

Using , t=k r k r we have sinkr

media and we have 1/ ak n c=

1 2sin sin tn n = .

Dont worry, we havent done all

A few comments.

Note that we can restateconserved on reflection a

As mentioned, Snells Laal-`Ala Ibn Sahl (940-10

also studied the mathema

design of burning mirror

centuries later in theBoo

a wave nature of light an

Sameen Ahmed Khan in

ne of the reflecting surface, so

al to the reflecting surface. Thus the three w

g surface.

matically.

nventional to measure angles from the norme both waves are in the same medium, k k=

le of reflection

sint tk r = . Now however, the wavevector

2d /tk n c= , so we have SnellsLaw of Re

that hard work just to prove these two laws!

oth laws in the fashion: The parallel compo

d transmission. Well make use of this for

w has recently been discovered to have been

0 C.E.), a mathematician in the Abassid cou

ics of lens design.He and co-workers were

s, described in his 984 book, On the Burnin

of Optics, Persian physicist Kamal al-Din al

a correct explanation of the rainbow. For d

ptics & Photonics News 18 (10), 22 (2007).

avevectors lie in the

al to the plane, so this

rk= krand so r = ,

s are in different

raction:

nent of wavevector is

later on.

known to Abu Sa`d

rt in Baghdad, who

nterested in the

g Instruments. Three

-Farisi was proposing

tails see the article by


13/25

13

Extension The figure below shows a plane wave incident on a boundary between two

media. The wavefronts are related by n11 =0= n22. Use this and the fact that thewavefronts must match up along the boundary to derive Snells Law.

2.2 Quantitative Reflection and Transmission at a Boundary: The FresnelEquations

At this point, weve established the directions of the reflected and refracted (transmitted) beams, but

we havent determined how much of the incident energy is transferred into each. It turns out that

this depends on thepolarisation of the incident light. While complicating the mathematics

somewhat, this has some very useful consequences.

Let the boundary between the two media be thex-y plane, and the plane of incidence, reflection and

refraction be the x-z plane. (There is no loss of generality in these choices).

Recall, that we can always decompose a plane wave into two orthogonally polarised plane waves.

There are two cases for reflection at a boundary depending on the orientation of the E field with

respect to the plane of incidence.

For the case where the incident wave Eis linearly polarised in the +y direction, that is the E field is

perpendicularto theplane of incidence, all of the following terms are used

Transverse Electric (TE) wave polarisation

s (senkrecht) polarisation

polarisation

For the case where the incident wave Eis linearly polarised in the plane of incidence, (or the B field

is perpendicular to the plane of incidence), all of the following terms are used

Transverse Magnetic (TM) wave.

polarisation

p (parallel) polarisation polarisation

Note that the orientation can be tricky to remember. For TE, the electric field isperpendicular tothe plane of incidence, butparallel to the interface.

t

n1, 1

n2, 2

i


14/25

14

Extensioncome up with theremembering which iss and

Lets begin with TE polarisation.

2.2.1 Reflection and Transmissi

Given that E, B, k are mutually

From Eq. (1.13), we know that th

across the boundary.

Thus, for TE, we have

,r tE E E B+ =

Recall that the magnitudes of the

1 1cos cosr rn E n E n

n

=

=

Rearranging this, we define the r

1 2

1 2

cos cos

cos cos

tr

t

n nEr

E n n

=

+

Where we have used =r. If we

( )

1 1

1 1

cos co

cos co

i i r

i t i

n E n E

n E n E E

we obtain the transmission coeffi

1

1 2

2 cos.

cos cos

t

t

E nt

E n n

=

+

Thus, for the TE case,1 2

1 2

cos cos,

cos cos

tTE

t

n nr

n n

=

+

n1

n2

most interesting and humorous mnemonhich isp. (There will be a prize).

..

on of TE Plane Waves (P&P 23-1)

rthogonal (and form a right hand set), we ca

e components ofE and Hparallel to a boun

os cos cos .r r t t B B =

fields satisfyB =E/v =E/(c/n), giving

( )

2

2

cos

cos

t t

r t

E

E E

+

eflection coefficient

,

instead eliminateErusingEr=Et-Ei in

2

2

s cos

s cos ,

r t t

r t t

n E

n E

=

=

cient

1

1 2

2 cos.

cos cosTE

t

nt

n n

=

+

Br

z

TEN

B c

ic you can for

draw the diagram:

ary are continuous

(1.15)

x

ki

kr Bi

Er

Ei

y

ormal Incidence

anges direction


15/25

15

Note that as a consequence ofE

1TE TE t r= + .

To use these formulae, we need

including Pedrotti3, use Snells L

incidence angle . This is entirel

2.2.2 Reflection and Transmissi

The argument runs similarly for

cos cos cor r tE E E + =

Again we useB =E/v = nE/c t

1 1 2r tn E n E n E + =

and solving these we get

2 1

2 1

cos cos,

cos cos

tTM

t

n nr

n n

+=

+

The TM coefficients satisfy

( )2 1 1TM TM n t n r = + .

2. 3 Notes on Fresnel EquLets make some observations ab

2.3.1 Sign conventions

The sign of the reflection coefficincoming and outgoing fields. In

which case the reflection coeffici

r tE E= the reflection and transmission coef

oth and t. We find the latter using Snells

aw to rewrite the reflection coefficients pure

y a matter of taste.

on of TM Plane Waves

M polarisation, but this time we have

s , .t r tB B B + =

o get

1

2 1

2 cos.

cos cos

TM

t

nt

n n

=

+.

tionsout properties of the Fresnel equations...

ents depends on our conventions for the orieTM, we could have chosen the B field to al

ent would differ by a minus sign. So in wor

n1

n2

z

TM

B

icients satisfy

Law. Some texts,

y in terms of the

(1.16)

ntation of theays lie along +y, in

ing out interference

k

kr

B

Br

ErE

y

ormal Incidence

hanges direction


16/25

16

effects etc, its important to keep the field orientations in mind. Conventions in other texts may

differ as well.

2.3.2 Power coefficients

The reflection coefficient ris an amplitude reflectivity. The power reflectivity or reflectance is

given by

2 *R r rr= = .

(In general, the refractive indices may be complex, so we use

the modulus squared). However, the powertransmission or

transmittanceTt2 except at normal incidence, because we aremoving from one medium to another. Therefore the beam area

changes, and the irradiance (power/unit area) changes too. We

can find the power transmission for a beam from

T= 1-R

using conservation of energy.

Extensioncan you find a different equation for the power transmission, by using theFresnel coefficient and Snells Law?

2.3.3 Simple examples

a) Glass-air interface at normal incidence (i = t= 0), n1 = 1, n2 = 1.5,

( ) ( )1 2 1 2/ 0.2Nr n n n n= + = (minus sign gives a 180 phase change reflecting from an optically

denser medium), so 0.04N NR r2= = (=4% reflectivity).

Note for Ge, n = 4, andRN= 36% per surface.

b) Power reflectivity (reflectance) at normal incidence from air:21

1

nR

n

= +

c) Sanity check, ifn1 = n2 (ie. no interface) then r= 0, t= 1 (good!)

2.3.4 The Brewster angle

We can obtain a very important result as follows. In the expression for the TM reflectivity we use

Snells law to eliminate2 1/n n and find

sin

cos cossin.

sincos cos

sin

tt

TM

t

t

r

+=

+

Now using the following rather obscure trigonometric identity

( )

( )

tan cos sin cos sin

tan cos sin cos sin

a b a a b b

a b a a b b

=

+ +,

we have

( )

( )

tan.

tan

t

TM

t

r

=

+

This equation shows that 0TMr = if / 2t + = . Now using Snells law, we can write

A1

A2


17/25

17

1 2 2 2sin sin sin( ) cos2

tn n n n

= = = ,

and dividing by cos yields the condition

2

1

tann

n = .

The angle ( )1 2 1tan / p n n= is known as Brewsters angle. With this condition there is zero

reflection forp polarisation. This means that if the incident beam is unpolarised, the reflected beam

is purely s polarised. This is the reason that polarising sunglasses are useful.

eg. Air to glass, n2 = 1.5, p = 56.3

At Brewsters angle, the reflected and transmitted beams are at right angles to each other, ie

/ 2r t t + = + = . This has a nice interpretation in terms of the following diagrams.

Extensionshow that the reflected and transmitted beams are at right angles.

It is common practice to use Brewster angled windows on the glass surfaces in lasers, so that there

are no reflection losses for one polarisation.

Note that there is a Brewsters angle for incidence from either side of the boundary, and the two

angles sum to 90.

Extension

Prove this last statement: the sum of the Brewsters angles from either side of theinterface is 90.

Why is there no Brewster angle for TE polarization? Try setting rTE=0 and find out

what goes wrong.

2.3.5 Total internal reflection

We can use Snells law to replace2 cos tn in the Fresnel equations by

2 2 212 1

2

( / ) sini

nn n

n but

for1 2sinn n > we have a complex reflectivity.

This can only happen when going from a medium of high index to a medium of lower index, ie. for

1 2n n> , and is known as total internal reflection (TIR). TIR occurs for angles of incidence greater

than the critical angle 2 1sin / c n n = and for both polarisations.

kr

Erinduces oscillating dipoles

which dont radiate inkr

direction

n1 n2

kr

ki

kt

Er

Ei

Et= 0

n1 n2


18/25

18

Note that the numerator and den

so as expected, we have2

R r=

By surrounding a high index me

region. This is the basic mechan

of the information revolution. In

cladding is tiny ( co cl1.444,n n

Below, we see that although the

is no optical energy in the low in

2.3.6 Phase change on reflectio

A negative value for the reflectio

the reflected electric field behave

phase shift depends on the polari

the Brewster angle:

Extension

Calculate BrewsA fisherman wehead approx 3far out from thefish easiest to s

Total internal re

Brewster anglerelation betweereverse?

Signed reflection coefficient for e

minator of the reflection coefficient are com

1 .

ium with a low one, we can effectively trap

sm of light guidance in the optical fibre and

silica optical fibres, the index difference bet

1.440 ). The critical angle is thus very clos

nergy is perfectly reflected at the interface, i

ex region.

n coefficient indicates a phase change upo

s as if it had been delayed by half a wavelen

ation, and for TM light, whether the angle o

ters angle for water (n = 1.334).ring Polaroidsunglasses stands on a rabove the level of the water in the river.

iver bank will the reflected glare be the lot assuming there are any)?

lection only occurs on one side of an int

xists for incidence from either side. Whthe Brewster angles from medium 1 to

xternal/internal reflection

plex conjugates and

ight in the high index

thus lies at the heart

een the core and

e to 90.

ts nottrue that there

reflection. That is,

th. The presence of a

incidence is beyond

iver bank with hisApproximately howast (and hence the

rface, but a

? What is theedium 2 and the


19/25

19

Reflectances for internal and external incidence.

Phase changes in TIR regime

In the TIR regime, the phase change upon reflection gets more interesting.

Consider the TE reflection coefficient,

1 2

1 2

cos cos

cos cos

cos cos ,cos cos

tTE

t

t

t

n nr

n n

nn

=

+

= +

Where2 1/ 1n n n= < . Using Snells Law, we can write

2 2 2 2 2

1 2cos 1 sin 1 ( / ) sin sin

t tn n n n n n = = = ,

to find2 2

2 2

2

cos sin

cos sin

e

TE

i

i nr

i n

=

+

=

with ( )2 2tan sin / cosn = . The total phase change on reflection is thus

2 21 sin2 2 tan

cosTE

n

= =

.

The exact formula is not so interesting. What does matter is that the phase change for total internal

reflection of TM light is different:

2 21

2

sin2tan

cosTM

n

n

= +

.

There are thus a number of phase change regimes, that depend on polarisation, external or internalincidence, and angle of incidence compared to the critical angle and Brewsters angle. The results

are summarised in these graphs:


20/25

20

Example

Find the phase shift for

internal and external

reflection of both

polarisations for a glass

(n=1.5) - air interface, atincidence angles of 30 and

60.

Application: The Fresnel rhomb

In the TIR regime, the difference

angle of incidence or the materia

Suppose we want to convert line

taking light with equal compone

difference in phase changes for g

rhomb is designed to allow two s

quarter wave plate but using isot

Extensionreview the text onbetween linear components p

2.3.7 Energy distribution in TIR

As mentioned above, while there

critical angle (the time-averaged

energy present there in the formThe transmitted wave is given by

in phase change between TE and TM is tuna

s. This allows to make useful devices:

rly polarised light to circularly polarised lig

ts ofs andp and delaying one component b

lass we find a 45 phase difference when uch reflections before the light exits. This p

ropic materials.

circularly polarised light and explain whoduces circular polarisation.

Frustrated and Attenuated Total Internal Re

is no energy transport into the low index me

Poynting vector into the medium vanishes), t

f an evanescent wave. We see this as follo

ble by adjusting the

t. We can do this by

/ 2 . Plotting the

53. A Fresnelovides us with a

a / 2 phase shift

lection

ium beyond the

here is still optical

s.


21/25

21

( )

(

0

0

exp

exp

t t t

t

i t

t i t

=

=

E E k r

E k r

From the earlier geometry,

( sin ,0, ct t tk x z= k r

where 2 /tk n c= . From Snellsthat the transmitted field has the

( )

(

( )

2 1 2

1

sin cos

0

( / )( / sin )

0

sin /

0

e e

e

e e

t t t t i xk zk

t

i x n c n n

i x n c t z

t

t

t

=

=

=

E E

E

E

So the electric field has a propag

character along the negativez-ax

It is interesting to examine the en

effort, (try it if you like!), this ca

( ) {2 2

0 , e sin / ztt c E x n = S r

Observe that when we time-aver

vanishes. So the wave exists insi

boundary, rather than across it.

This has a number of interesting

real and imaginary parts of the re

introduce an additional high inde

then tunnel across the low index

analogous to quantum mechanic

identical). By measuring how mthe refractive index difference be

FTIR has also recently found

application in some cool multi-to

screen technologies. See

www.cs.nyu.edu/~jhan for details

Attenuated Total Internal Reflect

leads to a reduction in the energy

part of the index of the new medi

The exponential decay of the fiel

damping of waves in lossy medi

conductor could only penetrate a

different phenomena.

In an evanescent wave, eenergy sloshes around lik

)os t ,

Law, sin sin / t n = , and cos sit t tk i k =orm

)e

.

i t

i i t

,

ting character along thex axis, and a decayi

is into the second medium.

ergy flux as described by the Poynting vecto

be shown to be

( ) (2 2 2cos sin sin / 1 sin 2k x t z n +

ge, the energy flow along thex direction sur

de the low index medium, but the average en

applications for laboratory devices, particula

fractive index. In Frustrated Total Internal

x medium in the region of the evanescent wa

medium and escape into the new one. This p

l tunnelling through a potential barrier (the

ch energy is transported into the third meditween the materials.

uch

:

ion is a similar idea except that the new med

in the reflected beam. Measuring reduction

um to be determined.

into the low index medium is reminiscent

which we saw in PHYS301. (Recall that an

small distance called the skin depth.) In fact

ergy is localised near the surface but there is

a standing wave.

2 2

n / 1n i = , so

g exponential

r. After considerable

) }sin / 2x t vives but that alongz

ergy flow is along the

ly for measuring the

eflection, we

ve. The wave can

rocess is exactly

athematics is

m we can measure

um is lossy which

allows the imaginary

f the exponential

EM wave in a

, these are very

no dampingthe


22/25

22

In a lossy medium, energy is removed from the wave and dissipated by doing work on thefree carriers (typically electrons). Any energy that enters the medium as light is doomed to

become heat.

2.5 Negative Index Materials and Left Handed Light

We have seen that the refractive index can be written:

0 0

r r

p

cn

v

= = =

Now it is quite common for to be negative (when the optical frequency is greater than anabsorption resonance frequency, or in metals when the optical frequency is greater than the plasma

frequency). Negative is also possible in resonant magnetic systems.

In an obscure 1968 paper, the Soviet physicist Veselago considered systems in which both and were negative and showed that this would imply a negative refractive index, with strange

consequences well examine below. However, since a material with both and simultaneouslynegative had never been discovered, this remained a theoretical curiosity. The reason that nature

apparently doesnt provide these materials is easy to understand. The negative permittivity and

permeability are found near resonances. Resonances for permittivity tend to be associated with

individual atomic and molecular electron states, and these occur in the ultraviolet and infrared. But

magnetism is associated with collective motion of many electrons, and so these resonances occur at

much lower frequencies, in the microwave and terahertz regimes. So the two types of resonances

just dont overlap in natural structures.

Then in the late 1990s, Sir John Pendry (then just John Pendry!), showed that by cleverlyengineering structures on a scale much smallerthan the wavelength, new resonances could be

introduced at virtually any frequency and so artificial materials could exhibit negative and . Asa result, the topic ofelectromagnetic metamaterials has exploded in the last decade. The cloaking

devices we examined in PHYS301 are examples of these kinds of systems.

A very early example of such a material is

shown at right (for operation in the

microwave). (Image from Andrew Houck,

Photonics Spectra). Observe that it consists of

a series ofsplit ring resonators and vertical

stripes. The former induce a negativemagnetic response and the latter a negative

electric response.

In the last couple of years, fabrication abilities

and theoretical understanding have exploded

in this area and there are now examples working (just barely!) in the optical. A few examples of

these recent structures are shown below.

Lets explore the consequences ofnegative materials.


23/25

23

Given0 0

n

= , if only one ofor is negative, then n becomes complex. This means that

( )i knz t e

includes a real exponential term, which corresponds to a non propagating wave. So such

materials are reflective and absorbing, and not terribly interesting.

In fact, to preserve causality (the physicists single most treasured principle), it turns out that both

and must be complex with positive imaginary parts. Now suppose their real parts are both

negative. Then ( )( )1/2

r r r i i r n i + + . Now, if [ ]Im 0n < , then waves would grow which is

impossible without a source of gain. We can then quickly show from the Argand diagram that

[ ]Re 0.n <

Restating, the energy requirement that the imaginary part ofn is positive, leads to the conclusion

that if both and have negative real parts, the real part ofn must also be negative.

One of Veselagos curious observations was that such materials obey Snells law, and therefore

transmit on the same side of the normal to the boundary.

From n1sin 1 = n2sin2, ifn1 is 1 (air) then 2

becomes negative for positive 1 :

Such materials can be used to make a perfect lens, ie. a lens for which there is no resolution limit,regardless of the wavelength. (This fact will be more surprising by the end of the course).

Complete the picture below to see how even a rectangular slab of metamaterial can form an image

of an object.

Simulation of wavefronts propagating frompositive n to negative n topositive n.

http://www.cmpgroup.ameslab.gov/portal/LHM.jpg/view

Object

1 2

n1 n2

Metamaterialn2 < 0


24/25

24

Experimental demonstrations of

in August 2008 by the group of

Nature and Science !).

Images from J. Valentine etnegative refractive index, N

More recently, simple examples

date they have not used negative

Images from T. Ergin et al, Wavelengths, Science 328,This paper creates an undetectabl

hidden:

egative index materials in the optical domai

iang Zhang at UC, Berkeley, (with simultan

l, Three-dimensional optical metamaature455, 376 (2008):

f optical cloaks in the optical have been de

index structures:

Three-Dimensional Invisibility Cloak at337 (2010):e depression in a boundary behind which an

n were first published

ous papers in both

erial with a

onstrated, though to

Optical

object could be


25/25

The images at left show the reflected light

from the uncloaked and cloaked

structures. The uncloaked structure

shows dim areas (green) where the light

has been reflected away by the bump.

The cloaked structure shows an almostflat response, indicating the depression is

hidden.

Phys 306 Lectures 1 Waves Media 2011

Documents